Measure, Integration & Real Analysis, 2021a

250 Chapter 8 Hilbert Spaces
Riesz Representation Theorem, Revisited
Now that we know that every Hilbert space has an orthonormal basis, we can give a
completely different proof of the Riesz Representation Theorem (8.47) than the proof
we gave earlier.
Note that the new proof below of the Riesz Representation Theorem gives the
formula 8.77 for h in terms of an orthonormal basis. One interesting feature of this
formula is that h is uniquely determined by ϕ and thus h does not depend upon the
choice of an orthonormal basis. Hence despite its appearance, the right side of 8.77
is independent of the choice of an orthonormal basis.
8.76 Riesz Representation Theorem
Suppose ϕ is a bounded linear functional on a Hilbert space V and {e k } k∈Γ is an
orthonormal basis of V. Let
8.77 h = ∑ ϕ(e k )e k .
k∈Γ
Then
8.78 ϕ( f )=〈 f , h〉
for all f ∈ V. Furthermore, ‖ϕ‖ = ( ∑ k∈Γ |ϕ(e k )| 2) 1/2 .
Proof First we must show that the sum defining h makes sense. To do this, suppose
Ω is a finite subset of Γ. Then
) ∥ (
∑ |ϕ(e j )| 2 ∥∥
= ϕ(
∑ ϕ(e j )e j ≤‖ϕ‖ ∥ ∑ ϕ(e j )e j = ‖ϕ‖ ∑ |ϕ(e j )| 2) 1/2
,
j∈Ω
j∈Ω
j∈Ω
j∈Ω
(
where the last equality follows from 8.52. Dividing by ∑ j∈Ω |ϕ(e j )| 2) 1/2
gives
(
∑ |ϕ(e j )| 2) 1/2
≤‖ϕ‖.
j∈Ω
Because the inequality above holds for every finite subset Ω of Γ, we conclude that
∑ |ϕ(e k )| 2 ≤‖ϕ‖ 2 .
k∈Γ
Thus the sum defining h makes sense (by 8.54) in equation 8.77.
Now 8.77 shows that 〈h, e j 〉 = ϕ(e j ) for each j ∈ Γ. Thus if f ∈ V then
( )
ϕ( f )=ϕ ∑ f , e k 〉e k
k∈Γ〈
= ∑
k∈Γ
〈 f , e k 〉ϕ(e k )=∑ 〈 f , e k 〉〈h, e k 〉 = 〈 f , h〉,
where the first and last equalities follow from 8.63 and the second equality follows
from the boundedness/continuity of ϕ. Thus 8.78 holds.
Finally, the Cauchy–Schwarz inequality, equation 8.78, and the equation ϕ(h) =
〈h, h〉 show that ‖ϕ‖ = ‖h‖ = ( ∑ k∈Γ |ϕ(e k )| 2) 1/2 .
k∈Γ
Measure, Integration & Real Analysis, by Sheldon Axler